Problem 140 - College Algebra Essentials

Question

Use the Pythagorean Theorem and the square root property to solve Exercises \(140-143 .\) Express answers in simplified radical form. Then find a decimal approximation to the nearest tenth. A rectangular park is 6 miles long and 3 miles wide. How long is a pedestrian route that runs diagonally across the park?

Step-by-Step Solution

Answer

The pedestrian route that runs diagonally across the park is \(3\sqrt{5}\) miles or approximately 6.7 miles.

Step 1: Identify the lengths of the other two sides

The two sides of the rectangular park are given as 3 miles and 6 miles.

Step 2: Apply Pythagorean Theorem

According to the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, it's \(c^2 = a^2 + b^2\), where c is the length of the hypotenuse. Substituting 3 for a and 6 for b, we get \(c^2 = 3^2 + 6^2\).

Step 3: Use square root property

After simplifying the right side of the equation above, to find the length of the hypotenuse, \(c\), you take the square root of both sides of the equation. So, \(c = \sqrt{9 + 36}\)

Step 4: Simplify the radical

Reduction of the expression inside the square root gives \(c = \sqrt{45}\), which simplifies to \(c = 3\sqrt{5}\) miles. This is the hypotenuse length in simplified radical form.

Step 5: Find the decimal approximation

To approximate the length of the hypotenuse to the nearest tenth, calculate the decimal form of \(3\sqrt{5}\), which will be 6.7 miles.

Key Concepts

Simplified Radical Form

When working with the Pythagorean Theorem, reaching the stage of a radical expression is quite common. A radical expression involves roots, most frequently square roots. In their raw form, these expressions can seem complex and unwieldy. Simplifying them is both an aesthetic choice and a practical necessity for further mathematical operations or comparisons.

To achieve a simplified radical form, we work to make the number under the radical as small as possible by factoring out perfect squares. Take, for example, \(\sqrt{45}\). This expression can be factored into \(\sqrt{9\times5}\). Since 9 is a perfect square, we can take its square root out of the radical, which is 3, leaving us with \(3\sqrt{5}\). This is now in simplified radical form—considerably neater and easier to work with, especially in equations or further mathematical procedures. It is important to note that simplification does not alter the value of the number; it simply presents it in its most reduced form.

Square Root Property

Square roots evoke the concept of undoing a square. If you square a number, say 4, you get 16. To return to the original number, you take the square root of 16. This is the essence of the square root property: if \(x^2 = y\), then \(x = \pm\sqrt{y}\). The \(\pm\) symbol is included because both the positive and negative values of \(x\) will yield \(y\) when squared. The positive value is known as the principal square root.

When applying this property to solve for the hypotenuse in a right triangle using the Pythagorean Theorem, we only consider the principal square root because a length cannot be negative. For our rectangular park example, to find the length of the diagonal, we took the principal square root of the sum of the squares of its sides, leading to \(\sqrt{45}\) or more simply, in its simplified form as \(3\sqrt{5}\) miles.

Decimal Approximation

In standard form, expressions can be quite exact but not always intuitive or convenient for everyday usage. This is where decimal approximation becomes handy. It allows us to express numbers, including those in radical form, as decimals which are more familiar and easily comparable.

The process of finding a decimal approximation typically involves a calculator, where the square root function can give us a number in decimal form. In our example, calculating the decimal approximation of \(3\sqrt{5}\) provides us with 6.7 miles. This value is easier to grasp and use in everyday contexts, like estimating the time it would take to walk the diagonal of the park. When approximating to the nearest tenth, we maintain a balance between accuracy and simplicity, making the result both useful and precise enough for practical situations.